Oxidation-Reduction Titrations
Dr M.Afroz Bakht
Introduction

Oxidation-Reduction (Redox): Definitions & Terms.

Oxidation Number Principles.

Balancing Redox Equation by Half-Reaction Method.
Electrochemical Cells and Electrode Potential.



Oxidation Potential: Definition and Factors Affecting.
Redox Titration Curves.

Detection of End point in Redox Titrations.
Standard Oxidizing Reagents and their Properties.

Applications of Redox Titrations.

Oxidation-Reduction (Redox)
Reaction of ferrous ion with chlorine gas
2Fe3+ + 2Cl
3 Fe2+ + Cl2
3 Fe2+ — 2e
Cl2
+ 2e
2 Fe3+ (Loss of electrons: Oxidation)
2 Cl-
(Gain of electrons: Reduction)
In every redox reaction, both reduction and oxidation must occur.
Substance that gives electrons is the reducing agent or reductant.
Substance that accepts electrons is the oxidizing agent or oxidant.
Overall, the number of electrons lost in the oxidation half reaction
must equal the number gained in the reduction half equation.
Oxidation Number (O.N)
 The O.N of a monatomic ion = its electrical charge.
 The O.N of atoms in free un-combined elements = zero
 The O.N of an element in a compound may be calculated by
assigning the O.N to the remaining elements of the
compound using the aforementioned basis and the
following additional rules:
 The O.N. for oxygen = –2 (in peroxides = –1).
 The O.N. for hydrogen = +1 (in special cases= —1).
 The algebraic sum of the positive and negative O.N. of the
atoms represented by the formula for the substance = zero.
 The algebraic sum of the positive and negative O.N. of the
atoms in a polyatomic ion = the charge of the ion.
Oxidation Numbers of Some Substances
Substance
Oxidation Numbers
NaCl
H2
NH3
H2O2
LiH
K2CrO4
SO42KClO3
Na = +1, Cl = —1
H = 0
N = —3,
H = +1
H = +1,
O = —1
Li = +1,
H = —1
K = +1,
Cr = +6,
O = —2, S = +6
K = +1,
Cl = +5,
O = —2
O = —2
Oxidation states of manganese and nitrogen in different species
For manganese
Mn
Mn2+
0
+2
Species
NH3
N2H4
O.N.
–3
–2
Species
O.N.
Mn3+
MnO2
+3
+4
MnO42 MnO4 
+6
+7
For nitrogen
NH2OH N2
–1
0
N2O
NO
HNO2
HNO3
+1
+2
+3
+5
Balancing Redox Reactions using HalfReaction Method
• Divide the equation into an oxidation half-reaction
and a reduction half-reaction
• Balance these
–
–
–
–
Balance the elements other than H and O
Balance the O by adding H2O
Balance the H by adding H+
Balance the charge by adding e-
• Multiply each half-reaction by an integer such that
the number of e- lost in one equals the number
gained in the other
• Combine the half-reactions and cancel
Balancing Redox Reactions Using the HalfReaction Method
Al
0
+
HCl
+1 -1
=
AlCl3
+3 -1
+
H2
0
Oxidation:
( Al = Al3+ + 3e- )  2 = 6 e-
Reduction:
( 2 H+ + 2e = H2 )  3 = 6 e-
Redox:
2 Al + 6 H+ = 2 Al3+ + 3 H2
MnO4 + C2O42 + H+
Mn2+ + CO2 + H2O
 Balance each half reaction:
MnO4 + 5é
C2O42
Mn2+
2 CO2 + 2é
 Use the number of moles so as to make the electrons gained in
one reaction equal those lost in the other one
2 MnO4 + 5 C2O42
2 Mn2+ + 10 CO2
 Balance oxygen atoms by adding water
2 MnO4 + 5 C2O42
2 Mn2+ + 10 CO2 + 8 H2O
 Balance hydrogen atoms by adding H+
2 MnO4 + 5 C2O42 + 16 H+
2 Mn2+ + 10 CO2 + 8 H2O
Electrochemical Cells
Electrochemical cells consist of electrodes immersed in electrolyte
solution and frequently connected by a salt bridge
Solution Pressure.
The tendency of the metal to dissolve in a
solution of its salt.
Ionic Pressure. The tendency of the metal cations to deposit
on its metal dipped into its solution.
 Cu/Cu2+ system: ionic pressure > solution pressure.
Cu2+ leaves the solution to deposit on Cu rod
 Zn/Zn2+ system: solution pressure > ionic pressure.
Zn metal tends to dissolve forming Zn2+ in solution.
The potential difference between the metal rod (electrode)
and the solution is known as electrode potential (E)
anode//cathode
Cu/CuSO4 // ZnSO4 /Zn
Nernest Equation for Electrode Potential (E)
Et =
Eo
RT
+
nF
log [Mn+]
Et = electrode potential at temperature t.
E = standard electrode potential (constant depend on the system)
R = gas constant
T = absolute Temp. (t°C + 273)
F = Faraday (96500 Coulombs)
loge = ln (natural logarithm = 2.303 log)
n= valency of the ion
[Mn+] = molar concentration of metal ions in solution
E25 °C =
Eo
0.0591
+
log [Mn+]
n
o
Standard Electrode Potential (E )
E25 °C = Eo + 0.0591 log [Mn+]
n
Eo is the electromotive force (emf) produced when a half cell
(consisting of the elements immersed in a molar solution of its
ions) is coupled with a standard hydrogen electrode (E = zero).
System
E° (volts)
System
E° (volts)
Li / Li+
K / K+
Mg/Mg2+
Al / Al3+
Zn / Zn2+
Fe / Fe2+
–3.03
–2.92
–2.37
–1.33
–0.76
–0.44
Cd/Cd2+
Sn / Sn2+
H2 (pt) / H+
Cu / Cu2+
Hg / Hg2+
Ag / Ag+
–0.40
–0.13
0.00
+0.34
+0.79
+0.80
Measurement of the Electrode Potential
By connecting to another electrode (galvanic cell), an electric current
will then flow from the electrode having —ve potential to that having
+ve potential (from Zn electrode to Cu electrode)
The emf of the current can then be measured.
The normal hydrogen electrode is used as a reference electrode.
Normal Hydrogen Electrode (NHE)
Consists of a piece of platinum
foil coated with platinum black
and immersed in a solution of
1 N HCl (with respect to H+).
H2 gas (at 1 atm. Pressure) is
passed. Platinum black layer
absorbs a large amount of H2
and can be considered as a bar
of hydrogen, it also catalyses
the half reaction:
2H+ + 2e  H2
Under these conditions:
H2 electrode potential = zero
o
Standard Oxidation Potential (E )
E25 °C =
Eo
0.0591
+
log [Oxidized] / [Reduced]
n
It is the e.m.f. produced when a half cell consisting of an inert
electrode (as platinum), dipped in a solution of equal
concentration of both the oxidized and reduced forms (such
as Fe3+ / Fe2+), is connected with a NHE
+0.339
Oxidation Potentials: Electrochemical Series
Factors Affecting Oxidation Potential
1. Common Ion
E25 °C =
Eo
0.0591
+
log [Oxid] /[Red]
n
 The potential of MnO4/Mn2+ varies with the ratio [MnO4]/[Mn2+].
 If ferrous is titrated with MnO4 in presence of Cl , chloride will
interfere by reaction with MnO4 and gives higher results.
Zimmermann’s Reagent (MnSO4, H3PO4 and H2SO4)
 MnSO4 has a common ion (Mn2+) with the reductant that
lowers the potential of MnO4/Mn2+ system:
 Phosphoric acid lowers the potential of Fe3+/Fe2+ system by
complexation with Fe3+ as [Fe(PO4)2]3.
 Sulphuric acid is used for acidification.
2. Effect of pH
E MnO4/Mn2+ =
Eo
0.0591
+
log
5
[MnO4][H+]
[Mn2+]
The oxidation potential of an oxidizing agent containing
oxygen increases by increasing acidity and vice versa.
Potassium permanganate:
MnO4 + 8H+ + 5e  Mn2+ + 4H2O
0.0591
o
E MnO4/Mn2+ = E +
log
5
Potassium dichromate:
E Cr2O72/Cr3+ = Eo +
[MnO4][H+]8
[Mn2+]
Cr2O72 + 14H+ +6e  2Cr3+ + 7H2O
0.0591
log
6
[Cr2O72][H+]14
[Cr3+]
3. Effect of Complexing Agents
Iodine:
I2 + 2e  2I
E I2/I
[I2]
0.0591
= Eo +
log
2
[I]2
E (I2/2I) system increases by the addition of HgCl2 since it
complexes with iodide ions.
Hg2+ + 4I  [HgI4]2
Ferric:
(low dissociation complex)
Fe3+ + e  Fe2+ E Fe3+/Fe2+
0.0591
o
=E +
1
log
[Fe3+]
[Fe2+]
• E (Fe3+/Fe2+) is reduced by the addition of F or PO43 due to the
formation of the stable complexes [FeF6]3 and [Fe(PO4)2]3
respectively. Thus, ferric ions, in presence of F or PO43 cannot
oxidize iodide although Eo(Fe3+/Fe2+) = 0.77 while Eo(I2/2I ) = 0.54.
4. Effect of Precipitating Agents
Ferricyanide:
[Fe(CN)6]3 + e  [Fe(CN)6]4
[[Fe(CN)6]3]
0.0591
E Ferri/Ferro = Eo +
log
1
[[Fe(CN)6]4]
Addition of Zn2+ salts which precipitates ferrocyanide:
[Fe(CN)6]4- + Zn2+  Zn2 [Fe(CN)6]
The oxidation potential of ferri/ferrocyanide system to oxidize iodide
to iodine, although the oxidation potential of I2/2I- system is higher.
4. Effect of Precipitating Agents
Copper:
2 Cu2+ + 4 I  2Cu2I2 (ppt) + I2
0.0591
o
E Cu2+/Cu+ = E +
1
log
[Cu2+]
[Cu+]
In this reaction Cu2+ oxidized I although:
EoCu2+/Cu+ = 0.16 and EoI2/2 I = 0.54.
Due to slight solubility of Cu2I2, the concentration of Cu+ is
strongly decreased and the ratio Cu2+/Cu+ is increased with a
consequent increase of the potential of Cu2+/Cu+ redox couple to
about + 0.86 V, thus becoming able to oxidize iodide into iodine.
Redox Titration Curve
It is the plot of potential (E, volts) versus the volume (mL)
of titrant.
Example: Titration of 100 ml 0.1 N Ferrous sulphate by 0.1 N
ceric sulphate.
Ce4+ + Fe2+ → Ce3+ + Fe3+
The change in potential during titration can be either
measured or calculated using Nernest equation as follows:
At 0.00 mL of Ce+4 added, the potential is due to iron system:
E = Eo + (0.0592/n) log([Ox]/[Red])= 0.77 V
No Ce+4 present; minimal, unknown [Fe+3]; thus, insufficient
information to calculate E
 After adding 10 mL Ce4+: E = 0.77 + 0.0591/1 log 10/90 = 0.71 V
 After adding 50 mL Ce4+: E = 0.77 + 0.0591/1 log 50/50 = 0.77 V
 After adding 90 mL Ce4+: E = 0.77 + 0.0591/1 log 90/10 = 0.82 V
 After adding 99 mL Ce4+: E = 0.77 + 0.0591/1 log 99/1 = 0.88 V
 A adding 99.9 mL Ce4+: E = 0.77 + 0.0591/1 log 99.9/0.1 = 0.94 V
 After adding 100 mL Ce4+: the two potentials are identical
 After adding 100 mL Ce4+ (end point): the two potentials are
identical
E = EO1 + 0.0591/1 log [Fe3+]/[Fe2+]= 0.77 V
E = EO2 + 0.0591/1 log [Ce4+]/[Ce3+]= 1.45 V
 Summation of two equations:
2 E = Eo1 + Eo2 + 0.0591 log
 At the end point:
[Fe3+][Ce4+]
[Fe2+][Ce3+]
Fe2+ = Ce4+ , and Fe3+ = Ce3+
2 E = E o1 + E o2
E = ( E o1 + E o 2 ) / 2
E = 0.77 + 1.45 / 2 = 1.10 V
Potential (V)
0
20 40 60 80 100 120 140
Ce4+ (mL)
Detection of End Point in Redox Titrations
1. Self Indicator (No Indicator)
When the titrant solution is coloured (KMnO4):
KMnO4 (violet) + Fe2+ + H+  Mn2+ (colourless) + Fe3+.
The disappearance of the violet colour of KMnO4 is due to its
reduction to the colourless Mn2+.
When all the reducing sample (Fe2+) has been oxidized
(equivalence point), the first drop excess of MnO4 colours the
solution a distinct pink.
2. External Indicator
In Titration of Fe2+ by Cr2O72
Cr2O72 + 3Fe2+ + 14H+  2Cr3+ + 3Fe3+ + 7H2O
The reaction proceeds until all Fe2+ is converted into Fe3+
Fe2+ + Ferricyanide (indicator)  Ferrous ferricyanide (blue)].
Fe 2+ + [Fe(CN)6]3 → Fe3[Fe(CN)6]2-.
•The end point is reached when the drop fails to give a blue
colouration with the indicator (on plate)
•Less accurate method and may lead to loss or contamination of
sample.
3. Internal Redox Indicator
Redox indicators are compounds which have different colours
in the oxidized and reduced forms.
Inox + n e
=
Inred
They change colour when the oxidation potential of the titrated
solution reaches a definite value:
E = E° + 0.0591/n log [InOX]/[Inred]
When [Inox] = [Inred] , E = E°
Indicator colours may be detected when: [Inox]/[Inred] = 1/10 or
10/1 hence, Indicator range:
E = E°In ± 0.0591/n
Diphenylamine
2
E° = 0.76, n = 2.
H
H
H
N
N
N
diphenylbenzidine (colourless)
+ 2H+ + 2e
diphenylamine (colourless)
E < 0.73 V
Range = 0.73 – 0.79 V.
H
H
+
E < 0.73 V, colourless (red.).
E > 0.79 V, bluish violet (ox.).
+ 2e
N
N
+
diphenylbenzidine (violet)
E > 0.79 V
Ferroin indicator (1,10-phenanthroline-ferrous chelate).
E° = 1.147, n = 1.
Range = 1.088 – 1.206 V.
N
E < 1.088 V, red (red.).
E > 1.206 V, pale blue (ox.).
N
N
Fe2+
N
3
Ferroin (red)
Fe3+
3
Ferrin (pale blue)
+ e
4. Irreversible Redox Indicators
Some highly coloured organic compounds that undergo
irreversible oxidation or reduction
Methyl Orange
(CH3)2N
N
SO3Na + Br2 + 2 H2O
N
Methyl orange
(CH3)2N
NO + ON
SO3Na + 4 HBr
In acid solutions, methyl orange is red.
Addition of strong oxidants (Br2) would destroy the indicator
and thus it changes irreversibly to pale yellow colour
Properties of Oxidizing Agents
1. Potassium permanganate (KMnO4)
2.
3.
4.
5.
Potassium dichromate (K2Cr2O7)
Iodine (I2)
Potassium iodate (KIO3)
Bromate-bromide mixture
1. Potassium permanganate (KMnO4)
Very strong oxidizing agent, not a primary standard, self indicator.
In acid medium:
MnO4 + 8H+ + 5e
Mn2+ + 4H2O
It can oxidize: oxalate, Fe2+, Ferrocyanide, As3+, H2O2, and NO2.
MnO42
In alkaline medium: MnO4 + e
In neutral medium:
4MnO4 + 2H2O

MnO2 + 4OH- + 3O2
Unstable
2. Potassium dichromate (K2Cr2O7)
It is a primary standard (highly pure and stable).
Cr2O72 + 14H+ +6e
2Cr3+ + 7H2O
Used for determination of Fe2+ (Cl does not interfere); ferroin
indicator.
3. Iodine (I2)
Solubility of iodine in water is very small.
Its aqueous solution has appreciable vapour pressure: Prepared in I
I2 + I
I3 (triiodide ion)
Iodine solution is standardized against a standard Na2S2O3
Iodimetry: Direct titration of reducing substances with iodine
The reducing substances (Eº < + 0.54 V) are directly titrated with
iodine.
Sn2+ + I2  Sn4+ + 2I
2S2O32 + I2  S4O62 + 2I
(Self indicator or starch as indicator)
Iodometry: Back titration of oxidizing substances
The oxidizing substance (Eº > + 0.54 V)is treated with excess
iodide salt:
2MnO4 + 10I + 16H+  5I2 + 2Mn2+ + 8H2O
Cr2O72 + 6I- + 14H+  2Cr3+ + 3I2 + 7H2O
The liberated Iodine is titrated with standard sodium
thiosulphate (starch as indicator)
4. Potassium iodate (KIO3)
It is strong oxidizing agent, highly pure, its solution is prepared
by direct weighing.
IO3 + 5I + 6H+  3I2 + 3H2O (in 0.1 N HCl) Eq.W = MW/5
IO3 + 2I2 + 6H+  5I+ + 3H2O (in 4-6 N HCl) Eq.W = MW/4
IO3 + 2I + 6H+  3I+ + 3H2O
Eq.W = MW/4
Andrew’s Reaction
Determination of iodide with potassium iodate in 4-6 N
HCl
(chloroform as indicator)
Starch can not be used.
Potassium iodate prepared in molar
5. Bromate-bromide mixture
Upon acidification of bromate/bromide mixture, bromine is
produced:
BrO3 + 5 Br + 6 H+  3 Br2 + 3 H2O
Used for the determination of phenol and primary aromatic
amines:
OH
OH
+ 3Br2
dark
Br
Br
+ 3HBr
Br
Phenol
2,4,6- Tribromophenol
The excess Br2 is determined:
Br2 + 2I  I2 + 2 Br
&
I2 + 2 Na2S2O3  Na2S4O6 + 2 I
Chloroform is added (dissolve TBP & indicator). Starch can be used
Applications of Redox
Titrations
1. Determination of Free Metallic Elements
Metallic iron
Dissolved in FeCl3 solution & the produced Fe2+ is titrated with MnO4
Fe + 2 FeCl3  3 FeCl2
(Zimmerman’s Reagent)
5 Fe2+ + MnO4 + 14 H+  5 Fe3+ + Mn2+ + 7 H2O
2. Determination of Halogen-Containing Compounds
 Iodine in iodine tincture
Direct titration with thiosulphate
I2 + 2 S2O32  2 I + S4 O62
 Bromine & chlorine
Back titration with thiosulphate, after treatment with excess KI.
Cl2 + 2I  l2 + 2Cl
I2 + 2 S2O32  2 I + S4 O62
3. Determination of Peroxides
 Hydrogen peroxide
Permanganatometrically. Direct titration with KMnO4.
5 H2O2 + 2 MnO4 + 6 H+  5 O2 + 2 Mn2+ + 8 H2O
Iodometry. Back titration with thiosulphate, after adding excess KI.
H2O2 + 2 I + 2 H+  I2 + 2 H2O
4. Determination of Anions
Oxalates and oxalic acid are strong reducing agents and can be
titrated with standard KMnO4 at 60 °C in the presence of dilute
sulphuric acid.
5C2O42- + 2MnO4- + 16H+  10CO2 + 2Mn2+ + 8H2O